Question

In a pseudo first order hydrolysis of ester in water the following results were obtained:

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds. (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

• A. $3.0 \times 10^{-2} s ^{-1}$
• B. $1.98 \times 10^{-2} s ^{-1}$
• C. $4 \times 10^{-3} s ^{-1}$
• D. $5.20 \times 10^{3} s ^{-1}$

Answer 1

Answer: (B)

$k=\frac{2.303}{t}log \frac{\left[R_{0}\right]}{\left[R\right]} where \left[R_{0}\right] =0.55 M $
$ At 30 sec k_{1}=\frac{2.303}{t}log \frac{0.55}{0.31}=1.91\times10^{-2}s^{-1} $
$ At 60 sec k_{2} =\frac{2.303}{60} log \frac{0.55}{0.17}=1.96\times10^{-20}s^{-1} $
$ At 90 sec k_{3} =\frac{2.303}{90 } log \frac{0.55}{0.085}=2.07\times10^{-2} s^{-1} $
$\frac{\left(1.91\times10^{-2}\right)+\left(1.96\times10^{-2}\right) +\left(2.07\times10^{-2}\right)}{3} $
$ = 1.98 \times10^{-2 }s^{-1}$